# Solving Leetcode Interviews in Seconds with AI: Maximum Performance of a Team


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1383" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given two integers n and k and two integer arrays speed and efficiency both of length n. There are n engineers numbered from 1 to n. speed[i] and efficiency[i] represent the speed and efficiency of the ith engineer respectively. Choose at most k different engineers out of the n engineers to form a team with the maximum performance. The performance of a team is the sum of its engineers' speeds multiplied by the minimum efficiency among its engineers. Return the maximum performance of this team. Since the answer can be a huge number, return it modulo 109 + 7.   Example 1:  Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 2 Output: 60 Explanation:  We have the maximum performance of the team by selecting engineer 2 (with speed=10 and efficiency=4) and engineer 5 (with speed=5 and efficiency=7). That is, performance = (10 + 5) * min(4, 7) = 60.  Example 2:  Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 3 Output: 68 Explanation: This is the same example as the first but k = 3. We can select engineer 1, engineer 2 and engineer 5 to get the maximum performance of the team. That is, performance = (2 + 10 + 5) * min(5, 4, 7) = 68.  Example 3:  Input: n = 6, speed = [2,10,3,1,5,8], efficiency = [5,4,3,9,7,2], k = 4 Output: 72    Constraints:  1 <= k <= n <= 105 speed.length == n efficiency.length == n 1 <= speed[i] <= 105 1 <= efficiency[i] <= 108  

	# Explanation
	Here's the solution to the problem:

*   **Sort by Efficiency:** Sort the engineers in descending order based on their efficiency. This allows us to iterate through potential minimum efficiency values in decreasing order.
*   **Maintain a Priority Queue (Min-Heap):**  Use a min-heap to keep track of the `k` engineers with the highest speeds seen so far, as we iterate through the sorted engineers. The heap will allow to maintain the `k` largest speeds, efficiently.
*   **Calculate Performance and Update Max:** For each engineer (in sorted order of efficiency), add their speed to the total speed. If the heap size exceeds `k`, remove the engineer with the smallest speed from the heap and subtract his speed from the total speed. Compute performance = (total speed) * (current engineer's efficiency). Keep track of the maximum performance seen so far.

*   **Time Complexity:** O(n log n), due to sorting and heap operations.
*   **Space Complexity:** O(n), to store the sorted engineer indices, and the heap in the worst case.

	
	# Code
	```python
	import heapq

def maxPerformance(n: int, speed: list[int], efficiency: list[int], k: int) -> int:
    """
    Calculates the maximum performance of a team of at most k engineers.

    Args:
        n: The number of engineers.
        speed: A list of integers representing the speed of each engineer.
        efficiency: A list of integers representing the efficiency of each engineer.
        k: The maximum number of engineers allowed in the team.

    Returns:
        The maximum performance of the team modulo 10^9 + 7.
    """

    engineers = sorted(range(n), key=lambda i: efficiency[i], reverse=True)
    heap = []  # min-heap to store speeds
    total_speed = 0
    max_performance = 0
    modulo = 10**9 + 7

    for i in engineers:
        current_speed = speed[i]
        current_efficiency = efficiency[i]

        total_speed += current_speed
        heapq.heappush(heap, current_speed)

        if len(heap) > k:
            total_speed -= heapq.heappop(heap)

        max_performance = max(max_performance, total_speed * current_efficiency)

    return max_performance % modulo
	```